Transcript Atomic Theory - Ms. Drury's Flipped Chemistry Classes

Video AP 2.1
Quantum Mechanical Model
Schrodinger, de Brogli, and Heisenberg
solved mathematical equations to describe
the behavior of e- in the H atom as being
both particle and wavelike, which lead to
the quantum mechanical model. The QMM
specifies that each e- has a specific
energy, however, they do not follow a
specific path. Instead, there are areas of
probable e- location, which are called
orbitals.
Quantum Mechanical Model
The scientists chose to study the hydrogen
e- with the lowest energy (ground state),
which they labeled 1s. They found that the
e- is moving but not necessarily in circles.
Heisenberg Uncertainty Principle
It’s impossible to know both the location of
an e- and its velocity (speed) at the same
time. It is more probable to find an e- near
the nucleus. The size of the 1s orbital is
described as the radius of a sphere that
encloses 90% of the total e- probability.
Quantum Mechanical Model
These calculations continued until they
could describe any e- from any element.
The first number for the electron represents
the row that the element can be found in.
This is the energy level.
 The letter represents the sublevel the
electron is in, based on area of the periodic
table. It will tell you the shape of the orbital.
(s is spherical, p is lobes and d has 2
lobes).
 The superscript represents the number of
electrons in that sublevel.

S p d orbitals
P orbitals
D orbitals
Check Your Understanding…
Give the row number and sublevel of
each of the following elements:
a. Fluorine
2p
b. Carbon 2p
c. Manganese 3d
d. Sodium 3s
e. Phosphorous 3p
Aufbau Principle
As protons are added one by one to the
nucleus to build up the elements, so are e-.
A new and more specific e- configuration
can be written using all the first three
quantum letters. Here is the order to fill the
orbitals:
1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6 5s2 4d10 5p6
6s2 4f14 5d10 6p6 7s2 5f14 6d10 7p6
Examples

The configuration for Be ends at 2s and
the second element in. S it ends at 2s2.
Write everything before 2s2:
 Be is 1s2 2s2

The configuration for Sulfur ends at 3p4.
 S is 1s2 2s2 2p6 3s2 3p4
Problems?

You will notice that it seems to skip
around a lot and this is because this is
the order of the periodic table. This
shows that 4s is in fact a lower energy
level than 3d and 4f is higher energy
than many other sublevels in energy
level five.
Orbital Energies
Using a
diagram
like the
one to
your left,
it is easy
to show
the way
e- fill the
orbitals.
Hund’s Rule
Notice that Carbon’s 6th arrow is in the
second p orbital. Hund’s Rule states the
lowest energy configuration is one
having the maximum number of
unpaired electrons allowed by Pauli
Principle in a particular set of
degenerate orbitals. They should have
parallel spins.
 In english? Put one up arrow in each
box before any get two.

Examples

How many unpaired electrons does
nitrogen have?

How many unpaired electrons does
nitrogen have?
Valence electrons
Valence e- are e- in the outermost principal
energy level. The rule still holds: the
elements in the same group have the
same number of valence e-. Therefore,
Nitrogen with a configuration of 1s22s22p3
and Phosphorous with a configuration of
1s22s22p63s23p3 both have 5 valence ebecause they are both in group 15.
Video AP 2.2
Ions

Remember, when ions from electrons
are added and subtracted to the valence
shell!
 Fluorine is 1s2 2s2 2p5
 The F- is 1s2 2s2 2p6
 Manganese is 1s2 2s2 2p6 3s2 3p6 4s2 3d5
 The Mn+2 ion is 1s2 2s2 2p6 3s2 3p6 3d5
Noble Gas Short Cut

Larger elements will have extremely
long configurations. A shortcut is to use
noble gas configurations. So Sodium
has 11e- and a configuration of
1s22s22p63s2 or [Ne]3s1.
Exceptions
*Half filled sublevels are not as stabled as filled sublevels,
but they are more stable than other configurations.
Ex.
Cr looks like it should be
1s22s22p63s23p64s23d4
But it is
1s22s22p63s23p64s13d5
Ex.
Cu looks like it should be
But it is
1s22s22p63s23p64s23d9
1s22s22p63s23p64s13d10
This rule is for all transition metals in groups 6 and 11.
Examples
Give the electron configurations using the
noble gas short cut for gold.
 [Xe] 6s2 5d9
 [Xe] 6s1 5d10
Isoelectronic

When two ions or atoms have the same
number of electrons. Example: Ar and
K+1.
 Argon:
 Potassium:
 Potassium Ion:
1s2 2s2 2p6 3s2 3p6
1s2 2s2 2p6 3s2 3p6 4s1
1s2 2s2 2p6 3s2 3p6
Magnetism
Paramagnetism: A type of induced
magnetism associated with unpaired
electrons that cause a substance to be
attracted to the inducing electric field.
 Diamagnetism: a type of magnetic field
associated with paired electrons that
cause a substance to be repelled from
the inducing electric plate.

M
a
g
n
e
t
I
s
m
Nodes



The diagram shows 1s, 2s, and 3s orbitals. The
colored areas are areas of high probability of finding
an e-. The areas that are white are areas of zero
probability of finding e-, which are called nodes.
The number of nodes and the size of the orbital
increase as the principal energy level increases. For s
orbitals the number of nodes equals n-1 where n is the
principle energy level.
P, d and f orbitals have a more complicated probability
distributions but it is important to remember that it is
more probable to find an e- near the nucleus.
Video AP 2.3
ER



ER is energy that
exhibits wave like
behavior and travels
through space at the
speed of light
(c = 3x108m/s)
Wavelength(λ):distance
between 2 peaks.
Frequency(v): waves per
second
Which wave is
more frequent?
Which has a longer
wavelength?
ER examples
1.
If the wavelength is 650nm, calculate the
frequency of light with units.
(3.0x108m/s)
2.
c=λv
=
(650x10-9m)(v)
v = 4.6x1014s-1
If the frequency is 200.s-1, calculate the
wavelength.
(3.0x108m/s)
= (λ)(200./s)
λ = 1.5x106m
Planck
Matter can transform into energy but only in
packets of energy called quantum. Planck
found that these energy packets are in
multiples of hv.
 ∆E=hv

h=Planck’s constant = 6.63x10-34Js
Planck
Example 3: CuCl in fireworks give off blue light with a
wavelength of 450nm. What is the amount of
energy emitted?
c = λv
(3.0x108 m/s) = (450x10-9m)(v)
v = 6.7x1014/s
E = hv
E = (6.626x10-34J/s)(6.7x1014/s)
E = 4.4x10-19J
Einstein
Einstein stated that ER or light is a stream of
particles called photons.
 Between Plank and Einstein, light is both
wave-like and particle-like.

E= hv = hc/λ
Because v=c/λ
deBroglie

If light is both wave and particle-like, can
matter be both as well?

deBroglie found electrons (which are
particles) are so small that they have a
measurable wavelength! So, YES!
λ=h/mυ
υ= velocity
Remember Light Spectra and Bohr?



Energy is released in quanta (packets) to produce
light.
When light is passed through a prism, colors may
be seen at various wavelengths.
Bohr measured the energy emitted to create his
quantum model of the atom.
Light Spectra and Bohr

En= -2.178x10-18J
n2
n= energy level
Atomics
1s
2s
2p 2p 2p
3s
3p 3p 3p
4s
3d 3d 3d 3d 3d
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O2-
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Na
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Na+
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Cl
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Cl-
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S
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S6+
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K
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K+
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Mn
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Mn2+
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Mn4+
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Mn7+
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Cu
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Cu+
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Cu2+
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Zn
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Fe
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